Express your answer in form a+bi (14+60i)+(-30+2i)
1 answer:
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For the first circle, let's use the pythagorean theorem
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.
so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.
when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.
now, let's check for second circle
well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.
let's check the third circle
this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.